sols1_14 - Mathematical Finance 1(a(1 0.1/2)2 = 1.1025 so...

Mathematical Finance1. (a) (1 + 0.1/2)2= 1.1025 so the effective rate is 0.1025.(b) (1 + 0.1/4)4-1≈0.1038.(c)e0.1-1≈0.1052.2. (a) 6.93 years(b) 28.41 years. Time to reachαDist= logα/log(1 +r).3.P(r) = 0 is equivalent to a quadratic inr. Solving this we findA= 300 has rate ofreturnr≈ -0.148,A= 500 hasr= 0,A= 700 hasr≈0.123.4. Answer:a= 1426.15.5. The present value of this sequence is-30.75<0 so it’s not worth investing.6. The present values of the first bond are (a) 1746.29(b) 90.26(c)-618.93. Thesecond bond has present value 40.94.7. The rate of return per month is thatrthat solves the equation117600 = 60036summationdisplayk=1(1+r)-k+120000(1+r)-36=6001 +r1-(1 +r)-36r/(1 +r)+120000(1+r)-36Answer: solve with Newton’s method or bisection to getr≈0.5616%.8. Group the positive terms and group the negative terms. Factor out exactly the rightpower of 1+r(key step). Show that what remains is monotone and takes both positiveand negative values. Deduce thatP(r) = 0 has a unique root.9. Due to the different interest rates you have to determine the balance after each year.

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